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3 years ago

0.825167 rounded to the nearest ten-thousandth

Mathematics

1 answer:

antoniya [11.8K]3 years ago

3 0

0.825200
(correct me if I’m wrong)

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City Community College plans to increase its enrollment capacity to keep up with an increasing number of student applicants. The

aivan3 [116]

The given above may be modeled by the arithmetic sequence with initial value (I) 2200 and common difference (d) of 70. The number of applicants every year can be written by the equation,
at = a1 + (n - 1) x d
From the given above, n is equal to 4. This corresponds to the term which is 3 years from now.
at = 2200 + (4 - 1) x 70 = 2410
Thus, the enrollment capacity would be 2410 students.

3 0

3 years ago

Can someone please help me

Rus_ich [418]

Answer:

1) A.

2) C.

Step-by-step explanation:

1) 1 pint = 1/2 quarts

15 pints = 15 × 1/2 quarts

15 pints = 15/2 quarts

15 pints = 7 1/2 quarts

2) 1 quarts = 0.95 liters

4 quarts = 0.95 × 4 liters

4 quarts = 3.8 liters

5 0

3 years ago

Again, first answer is indeed, factual but please help.

ollegr [7]

Answer:

Part B: 4 is the constant

Part C: y = 4x

Step-by-step explanation:

y = kx

k = constant

y = 4x

4 is the constant

7 0

3 years ago

Which of the following is the solution to the differential equation....

bixtya [17]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}= (x-2)(y-2)\\ y>2\\\frac{dy}{y-2}=(x-2)dx\\$

integrate both side

<h2>∫ $\frac{dy}{y-2}=$ ∫ $(x-2)dx\\$ </h2><h2> $ln(y-2)=\frac{(x-2)^2}{2} +C\\$ </h2><h3>Substitute y(4)=5</h3><h3> $ln3=2+C\\C=ln3-2$ </h3><h3> $ln(y-2)=\frac{(x-2)^2}{2} +ln(3)-2\\y-2=e^{\frac{(x-2)^2}{2} +ln(3)-2}\\y=2+3e^{\frac{x^2}{2}-2x }$ </h3><h3 />

4 0

3 years ago

Can someone please turn this into an equation? Also would the equation be substitution or elimination?

irina1246 [14]

Answer:

A; C=600 B: C=1.25s+359

Step-by-step explanation:

7 0

3 years ago